How do you solve the inequality (x^2-2x-24)/(x^2-8x-20)>=0?

Apr 1, 2018

The solution is $x \in \left(- \infty , - 4\right] \cup \left(- 2 , 6\right] \cup \left(10 , + \infty\right)$

Explanation:

Factorise the inequality

$\frac{{x}^{2} - 2 x - 24}{{x}^{2} - 8 x - 20} = \frac{\left(x + 4\right) \left(x - 6\right)}{\left(x + 2\right) \left(x - 10\right)} \ge 0$

Let $f \left(x\right) = \frac{\left(x + 4\right) \left(x - 6\right)}{\left(x + 2\right) \left(x - 10\right)}$

Perform a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a}$$6$$\textcolor{w h i t e}{a a a a}$$10$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 6$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 10$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 4\right] \cup \left(- 2 , 6\right] \cup \left(10 , + \infty\right)$

graph{(x^2-2x-24)/(x^2-8x-20) [-25.65, 25.66, -12.83, 12.84]}